Is cos(x)*cot(x)+sin(x)=csc(x) ?

The answer to whether cos(x)*cot(x)+sin(x)=csc(x) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove cos(x)*cot(x)+sin(x)=csc(x)

prove

cos (x) \cdot cot (x) + sin (x) = csc (x)

True

Solution steps

cos (x) cot (x) + sin (x) = csc (x)

Manipulating left side

cos (x) cot (x) + sin (x)

Express with sin, cos

sin (x) + cos (x) cot (x)

Use the basic trigonometric identity:

cot (x) = \frac{cos ( x )}{sin ( x )}

= sin (x) + cos (x) \frac{cos ( x )}{sin ( x )}

Simplify

sin (x) + cos (x) \frac{cos ( x )}{sin ( x )} : \frac{sin ^{2} ( x ) + cos ^{2} ( x )}{sin ( x )}

sin (x) + cos (x) \frac{cos ( x )}{sin ( x )}

cos (x) \frac{cos ( x )}{sin ( x )} = \frac{cos ^{2} ( x )}{sin ( x )}

cos (x) \frac{cos ( x )}{sin ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{cos ( x ) cos ( x )}{sin ( x )}

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

= \frac{cos ^{2} ( x )}{sin ( x )}

= sin (x) + \frac{cos ^{2} ( x )}{sin ( x )}

Convert element to fraction:

sin (x) = \frac{sin ( x ) sin ( x )}{sin ( x )}

= \frac{sin ( x ) sin ( x )}{sin ( x )} + \frac{cos ^{2} ( x )}{sin ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( x ) sin ( x ) + cos ^{2} ( x )}{sin ( x )}

sin (x) sin (x) + cos^{2} (x) = sin^{2} (x) + cos^{2} (x)

sin (x) sin (x) + cos^{2} (x)

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

= sin^{2} (x) + cos^{2} (x)

= \frac{sin ^{2} ( x ) + cos ^{2} ( x )}{sin ( x )}

= \frac{sin ^{2} ( x ) + cos ^{2} ( x )}{sin ( x )}

= \frac{cos ^{2} ( x ) + sin ^{2} ( x )}{sin ( x )}

Rewrite using trig identities

\frac{cos ^{2} ( x ) + sin ^{2} ( x )}{sin ( x )}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= \frac{1}{sin ( x )}

= \frac{1}{sin ( x )}

Rewrite using trig identities

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

\frac{1}{\frac{1}{c s c ( x )}}

Simplify

\frac{1}{\frac{1}{c s c ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ( x )}{1}

Apply rule

\frac{a}{1} = a

= csc (x)

csc (x)

csc (x)

We showed that the two sides could take the same form

\Rightarrow True

csc (t) - sin (t) = cot (t) cos (t)

\frac{sec ( x ) + 1}{sec ( x )} = 1 + cos (x)

\frac{cos ( A )}{1 + sin ( A )} + tan (A) = sec (A)

- 3 cos^{2} (θ) + 11 = - 5 sin (θ) + 6

(1 - tan (x))^{2} = 1 - 2 tan (x) + tan^{2} (x)

Is cos(x)*cot(x)+sin(x)=csc(x) ?
The answer to whether cos(x)*cot(x)+sin(x)=csc(x) is True